Question

32. The magnetic induction in air at a distance d from an isolated point pole of strength m unit will be

(a) $\frac{m}{d}$ (b) $\frac{m}{d^2}$ (c) md (d) md²

​33. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque required to maintain the needle in this position will be

Answer 1

Dear Student

Answer 32.
$$\begin{gathered} MagneticinductionB=\frac{magneticstrength}{Squareofdis\tan ce} \\ B=\frac{m}{d^2} \\ HenceOptionBisRight \end{gathered}$$


Answer 33.
M and B are parallel initially
to move from zero to 60^0.
The work done $$\begin{gathered} W=∆U=M\times B\left[\cos \right(0)-\cos (60\left)\right]=\frac{MB}{2} \\ W=\frac{MB}{2} \\ MB=2W \\ Letstorque=M\times B=MB\sin \theta =MB\sin \left({60}^0\right)=\frac{\sqrt{3}}{2}\left(2W\right)=\sqrt{3}W \\ Torque=\sqrt{3}W \end{gathered}$$
Regards